已知x=1/2(5^1/n-5^-1/n)(n属于正整数)求(x-[1+x^2])的值.注:[ ]代表根号,故1+x^2是根号里面的数

令a=5^(1/n)
则x=(a-1/a)/2
x-(1+x^2)^(1/2)
=x-[1+(a^2-2+1/a^2)/4](1/2)
=x-[(a^2+2+1/a^2)/4](1/2)
=(a-1/a)/2-(a+1/a)/2
=-1/a
=-5^(-1/n)

令a=1\n
x=0.5(5^a-5^-a)
=0.5(5^a- 1/ 5^a )
=0.5(5^2a/5^a - 1/ 5^a ) =0.5(5^2a-1/ 5^a)=5^2a-1 / 2*5^a
所以1+x^2=1+ (5^2a-1 / 2*5^a)^2
=1+ (5^4a- 2*5^2a+1 / 4*5^2a)
=5^4a-2*5^2a+1+4*5^2a/4*5^2a
=5^4a+2*5^2a+1/4*5^2a
=(5^2a+1)^2/(2*5^a)^2 这里已经出现了2个平方项
因为n是正整数 所以a也大于0 所以 5^2a+1 2*5^a都大于零
开方就是 5^2a+1/2*5^a =5^a/2 =5^1/n /2

-1/5